ChemistryWORKBOOK SACE STAGE 2Australian Curriculum

SECOND EDITION Rhys Lewis

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CONTENTS 3

ContentsABOUT THIS BOOK 4

CHAPTER 1 TOPIC 1: MONITORING THE ENVIRONMENT 5 1.1: Global warming and climate change 6 1.2: Photochemical smog 33

1.3: Volumetric analysis 42

1.4: Chromatography 87

1.5: Atomic spectroscopy 110

Review Test 1 130

CHAPTER 2 TOPIC 2: MANAGING CHEMICAL PROCESSES 139 2.1: Rates of reactions 140

2.2: Equilibrium and yield 161

2.3: Optimising production 192

Review Test 2 210

CHAPTER 3 TOPIC 3: ORGANIC AND BIOLOGICAL CHEMISTRY 219 3.1: Introduction 220

3.2: Alcohols 227

3.3: Aldehydes and ketones 242

3.4: Carbohydrates 258

3.5: Carboxylic acids 268

3.6: Amines 282

3.7: Esters 293

3.8: Amides 307

3.9: Triglycerides 314

3.10: Proteins 333

Review Test 3 345

CHAPTER 4 TOPIC 4: MANAGING RESOURCES 355 4.1: Energy 356

4.2: Water 399

4.3: Soil 415

4.4: Materials 439

Review Test 4 485

SOLUTIONS TO CHAPTER QUESTIONS AND REVIEW TESTS 493

PERIODIC TABLE OF ELEMENTS 587

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37CHAPTER 1 • TOPIC 1: MONITORING THE ENVIRONMENT

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2NO + 2CO → N2 + 2CO2

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38 SASTA STAGE 2 • CHEMISTRY WORKBOOK

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42 SASTA STAGE 2 • CHEMISTRY WORKBOOK

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𝜌𝜌 𝑐𝑐

𝜌𝜌 = 𝑚𝑚𝑉𝑉

𝜌𝜌𝑚𝑚V

𝜌𝜌 =𝑚𝑚V

𝜌𝜌 = 57.50.6𝜌𝜌 = 95.8 g. L−1

𝑚𝑚 = 𝜌𝜌𝑉𝑉

𝑚𝑚 = 60.0 x 0.25

𝑚𝑚 = 15.0 g

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54 SASTA STAGE 2 • CHEMISTRY WORKBOOK

C3H8 + 5O2 → 3CO2 + 4H2O

𝑛𝑛C3H8 =𝑚𝑚M 𝑛𝑛O2 =

𝑚𝑚M

𝑛𝑛C3H8 =50

44.094𝑛𝑛O2 =

12532.00

𝑛𝑛C3H8 = 1.1 mol (1.13) 𝑛𝑛O2 = 3.9 mol

𝑛𝑛CO2)

Ratio = 3𝑛𝑛CO2: 1𝑛𝑛C3H8 Ratio = 3𝑛𝑛CO2: 5𝑛𝑛O2

𝑛𝑛CO2 = 𝑛𝑛C3H8x 3 𝑛𝑛CO2 =𝑛𝑛O2

5 x 3

𝑛𝑛CO2 = 3.4 mol 𝑛𝑛CO2 = 2.3 mol (2.34)

𝑛𝑛CO2 = 𝑛𝑛O2(limiting reagent)

𝑚𝑚CO2 = 𝑛𝑛CO2MCO2𝑚𝑚CO2 = 2.34 x 44.01

𝑚𝑚CO2 = 103 g

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90 SASTA STAGE 2 • CHEMISTRY WORKBOOK

• 𝑅𝑅𝑓𝑓

𝑅𝑅𝑓𝑓

𝑅𝑅𝑓𝑓 =distance travelled by component

solvent front

𝑅𝑅𝑓𝑓 =A

solvent front𝑅𝑅𝑓𝑓 =

Bsolvent front

𝑅𝑅𝑓𝑓 =C

solvent front

𝑅𝑅𝑓𝑓 =1.58.0

𝑅𝑅𝑓𝑓 =4.18.0

𝑅𝑅𝑓𝑓 =6.88.0

𝑅𝑅𝑓𝑓 = 0.19 𝑅𝑅𝑓𝑓 = 0.51 𝑅𝑅𝑓𝑓 = 0.85

𝑅𝑅𝑓𝑓𝑅𝑅𝑓𝑓

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105CHAPTER 1 • TOPIC 1: MONITORING THE ENVIRONMENT

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134 SASTA STAGE 2 • CHEMISTRY WORKBOOK

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135CHAPTER 1 • TOPIC 1: MONITORING THE ENVIRONMENT

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202 SASTA STAGE 2 • CHEMISTRY WORKBOOK

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228 SASTA STAGE 2 • CHEMISTRY WORKBOOK

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229CHAPTER 3 • TOPIC 3: ORGANIC AND BIOLOGICAL CHEMISTRY

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237CHAPTER 3 • TOPIC 3: ORGANIC AND BIOLOGICAL CHEMISTRY

Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O

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277CHAPTER 3 • TOPIC 3: ORGANIC AND BIOLOGICAL CHEMISTRY

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293CHAPTER 3 • TOPIC 3: ORGANIC AND BIOLOGICAL CHEMISTRY

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325CHAPTER 3 • TOPIC 3: ORGANIC AND BIOLOGICAL CHEMISTRY

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356 SASTA STAGE 2 • CHEMISTRY WORKBOOK

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6CO2(g) + 6H2O(𝑙𝑙) → C6H12O6 (s) + 6O2 (g) ∆𝐻𝐻 = +2803 kJ. mol−1

C6H12O6 (s) + 6O2 (g) → 6CO2(g) + 6H2O(𝑙𝑙) ∆𝐻𝐻 = −2803 kJ. mol−1

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385CHAPTER 4 • TOPIC 4: MANAGING RESOURCES

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387CHAPTER 4 • TOPIC 4: MANAGING RESOURCES

H2 → 2H+ + 2e−

O2 + 4H+ + 4e− → 2H2O

O2 + 2H2 → 2H2O ∆𝐻𝐻 = −286 kJ. mol−1

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V2+ → V3+ + e−

VO2+ VO2+

VO2+ + 2H+ + e− → VO2+ + H2O

V3+ + e− → V2+

VO2+ + H2O → VO2+ + 2H+ + e−

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401CHAPTER 4 • TOPIC 4: MANAGING RESOURCES

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411CHAPTER 4 • TOPIC 4: MANAGING RESOURCES

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415CHAPTER 4 • TOPIC 4: MANAGING RESOURCES

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465CHAPTER 4 • TOPIC 4: MANAGING RESOURCES

Al3+ + 3e− → Al

2O2− → O2 + 4e−

C + O2 → CO2

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481CHAPTER 4 • TOPIC 4: MANAGING RESOURCES

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482 SASTA STAGE 2 • CHEMISTRY WORKBOOK

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SOLUTIONS TO CHAPTER QUESTIONS

AND REVIEW TESTSsam

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502 SASTA STAGE 2 • CHEMISTRY WORKBOOK

𝑛𝑛PbS =𝑚𝑚PbS

M

𝑛𝑛PbS =50 000

(207.2 + 32.06)𝑛𝑛PbS = 209 mol (208.98)2𝑛𝑛PbO : 2𝑛𝑛PbS𝑛𝑛PbO = 𝑛𝑛PbS𝑛𝑛PbO = 209 mol (208.98)

𝑛𝑛O2 =𝑚𝑚O2

M

𝑛𝑛O2 =15 000

(16.00 x 2)𝑛𝑛O2 = 469 mol (468.75)

2𝑛𝑛PbO : 3𝑛𝑛O2

𝑛𝑛PbO =𝑛𝑛O2

3 x 2

𝑛𝑛PbO =468.75

3 x 2

𝑛𝑛PbO = 313 mol (312.5)𝑛𝑛PbS𝑚𝑚PbO = 𝑛𝑛PbO x 𝑀𝑀

𝑚𝑚PbO = 208.98 x (207.9 + 16.00)𝑚𝑚PbO = 47 x 103 g (46 790)

𝑛𝑛PbO =𝑚𝑚PbO

M

𝑛𝑛PbO =45 000

(207.9 + 16.00)𝑛𝑛PbO = 201 mol (200.98)1𝑛𝑛Pb : 1𝑛𝑛PbO𝑛𝑛Pb = 201 mol (200.98)𝑚𝑚Pb = 𝑛𝑛Pb𝑚𝑚Pb = 200.98 x 207.9𝑚𝑚Pb = 42 x 103 g (41784)

3𝑛𝑛Pb : 2𝑛𝑛PbO

𝑛𝑛Pb =200.98

2 x 3

𝑛𝑛Pb = 302 mol (301.47)

𝑚𝑚Pb = 𝑛𝑛Pb 𝑀𝑀

𝑚𝑚Pb = 301.47 x 207.9

𝑚𝑚Pb = 6.3 x 104 g (62676)

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𝑅𝑅𝑓𝑓

𝑅𝑅𝑓𝑓

𝑅𝑅𝑓𝑓

𝑅𝑅𝑓𝑓 (B) =B

solvent front =1.4 cm3.6 cm

𝑅𝑅𝑓𝑓 (B) = 0.39

𝑅𝑅𝑓𝑓 (C) =C

solvent front =2.4 cm3.6 cm

𝑅𝑅𝑓𝑓 (C) = 0.67∆𝑅𝑅𝑓𝑓 = 𝑅𝑅𝑓𝑓 (C) − 𝑅𝑅𝑓𝑓 (B)∆𝑅𝑅𝑓𝑓 = 0.67 − 0.39∆𝑅𝑅𝑓𝑓 = 0.28

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